Problem: Find $\tan\left(\dfrac{19\pi}{12}\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3-\sqrt{6}}{3+\sqrt{6}}$ (Choice B) B $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}$ (Choice C) C $\dfrac{1+\sqrt{3}}{1-\sqrt{3}}$ (Choice D) D $\dfrac{1+\sqrt{3}}{2}$
Answer: The strategy First, we should rewrite the given angle $\dfrac{19\pi}{12}$ as the sum or difference of two special angles. Then, we can use the tangent addition or subtraction identities in order to evaluate $\tan\left(\dfrac{19\pi}{12}\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $\dfrac{19\pi}{12}$ We can rewrite $\dfrac{19\pi}{12}$ as follows. $\begin{aligned}\dfrac{19\pi}{12}&=\dfrac{15\pi}{12}+\dfrac{4\pi}{12}\\\\\\ &=\dfrac{5\pi}{4}+\dfrac{\pi}{3}\end{aligned}$ In other words, $\dfrac{19\pi}{12}$ is the sum of the special angles $\dfrac{5\pi}{4}$ and $\dfrac{\pi}{3}$. Evaluating $\tan\left(\dfrac{19\pi}{12}\right)$ Using the tangent addition identity, we get the following. $\begin{aligned} \tan\left(\dfrac{19\pi}{12}\right)&= \tan\left(\dfrac{5\pi}{4}+\dfrac{\pi}{3}\right) \\\\\\ &= \dfrac{\tan\left(\dfrac{5\pi}{4}\right)+\tan\left(\dfrac{\pi}{3}\right)}{1-\tan\left(\dfrac{5\pi}{4}\right)\tan\left(\dfrac{\pi}{3}\right)}\\\\\\ &= \dfrac{\left(1\right)+\left({\sqrt{3}}\right)}{1-\left(1\right)\left({\sqrt{3}}\right)}\\\\\\ &=\dfrac{1+\sqrt{3}}{1-\sqrt{3}} \end{aligned}$ Summary $\tan\left(\dfrac{19\pi}{12}\right) = \dfrac{1+\sqrt{3}}{1-\sqrt{3}}$